5.2 Definitions and preliminaries
5.2.4 Twisted Yangians
Twisted Yangian Y+(3) Let us now define the twisted Yangian Y+(3) [46]. In this paragraph we assume that i, j run through the set {−1,0,1}. Let us then introduce the S-matrix (5.5), which now takes the explicit form
S(u) :=T(u)Tt(−u), (5.32)
or in terms of its matrix elementsS(u) = (sij(u)) sij(u) =X
a
tia(u)t−j,−a(−u), (5.33)
wheretij(u)’s are the generators ofY(3). The twisted Yangian Y+(3) is generated bysij(u), therefore Y+(3) is a sub-algebra of Y(3).
Using (5.32), one can derive that the matrixS(u) satisfies the quaternary relation (5.6) and additionally the symmetry relation
St(−u) =S(u) + 1
2u(S(u)−S(−u)), (5.34)
From the symmetry relation (5.34) we can then obtain that the elements
s(k)11, s(k)10, s(k)01, s(2k)00 , s(2k)1,−1, s(2k)−1,1, k= 1,2, . . . , (5.35) constitute a system of linearly independent generators.
TheY+(3) moduleV is highest weight if there exists a nonzero vectorv∈V such that V is generated byv and
sij(u)·v = 0, fori < j, (5.36) sii(u)·v = µi(u)v, fori= 0,1. (5.37) In [46] it was shown that every finite dimensional irrep of Y+(3) is a highest weight representation (Theorem 3.3).
The mapping
Fij →s(1)ij , (5.38)
defines an inclusion U(so3) → Y+(3). We can choose the following conventions for the so3 ∼=sl2 generators
Sz=−F11=F−1,−1, (5.39)
S+=F01=−F−1,0, (5.40)
S−=F10=F0,−1. (5.41)
Using the defining equation (5.32) ofY+(3), thegl3 module L(λ1, λ2, λ3) defines also an Y+(3) module. Let |a1, a2, a3i ∈L(α, β, γ) such that
Eii· |a1, a2, a3i=ai|a1, a2, a3i. (5.42) Using (5.38) and (5.39)-(5.41), we can obtain that
Sz· |a1, a2, a3i= (a1−a3)|a1, a2, a3i. (5.43) We can also see that
• thegl3 generators E1,0 and E0,−1 decrease the sl2 weight by one,
• thegl3 generators E0,1 and E−1,0 increase the sl2 weight by one,
• thegl3 generator E1,−1 decreases thesl2 weight by two,
• thegl3 generators E−1,1 increases the sl2 weight by two.
In the following we will use the gl3 module L(λ1, λ1, λ2) which is a highest weight rep of Y+(3) with highest weight
µ1(u) = (1 +λ2u−1)(1−λ1u−1), (5.44) µ0(u) = (1 +λ1u−1)(1−λ1u−1). (5.45) Using Lax-operator we can write
sij(u)·v=X
a
L(λi,a1,λ1,λ2)(u)L(λi,−a1,λ1,λ2)(−u)v, (5.46) for all v∈L(λ1, λ1, λ2).
From the co-product of Y(3) (5.23), one can show that the twisted YangianY+(3) is a co-ideal sub-algebra ofY(3) i.e.
∆ (sij(u)) =X
ab
tia(u)t−j,−b(−u)⊗sab(u) ∈Y(3)⊗Y+(3). (5.47) Using this equation, any tensor productL⊗V of a Y(3) moduleL and aY+(3) module V is aY+(3) representation i.e.
y·(v⊗w) = ∆ (y) (v⊗w), (5.48) where y∈Y+(3), v∈Land w∈V.
Extended twisted Yangian X(so6,so5) Let us again introduce the matrix S-matrix (5.5) by setting
S(u) :=T(u)K(u)Tt(−u), (5.49)
where T(u) is the generating function of X(so6) and the K-matrix is explicitly given by
K(u) =
u
u+1 0 0 0 0 0
0 u+1u 0 0 0 0
0 0 −u+11 1 0 0
0 0 1 −u+11 0 0
0 0 0 0 u+1u 0
0 0 0 0 0 u+1u
. (5.50)
The algebra generated bysij(u) is the extended twisted YangianX(so6,so5) [48], therefore X(so6,so5) is a sub-algebra of X(so6).
The K-matrix satisfies the twisted Yang-Baxter equation (5.4) and the following symmetry equation
Kt(u) =K(−u)− 2u
(u+ 1) (u−1)1. (5.51)
Using (5.21) and (5.51) one can then derive that the matrix S(u) satisfies the following quaternary relation and symmetry relation
R12(u−v)S1(u)R12(u+v+ 2)S2(v) =S2(v)R12(u+v+ 2)S1(u)R12(u−v), (5.52) St(u) =S(−u) + 1
2u(S(u)−S(−u))− 1
2u−2tr (S(u))I. (5.53)
In [48] it was shown that the reflection algebras generated by (5.52) and (5.53) are isomor-phic toX(so6,so5) (Theorem 4.2). Unfortunately, the representation theory ofX(so6,so5) has hardly been studied in the literature. In [49,50] only twisted Yangians with diagonal K-matrix were studied. Therefore the twisted Yangian X(so6,so5) was ignored (see the explicit form of the K-matrix (5.50)). To the best of our knowledge the proper definition of theX(so6,so5) highest weight representations has not yet appeared in the literature. Nev-ertheless, using the algebra homomorphism (5.28) and investigating the algebra embedding U(so5)⊂X(so6,so5) we can conjecture that the following is the correct definition for the X(so6,so5) highest weight representations. The X(so6,so5) module V is highest weight if there exists a nonzero vector v∈V such thatV is generated by v and
sij(u)·v = 0, for all i < j where (i, j)6= (−1,1), (5.54)
sii(u)·v =µi(u)v, (5.55)
s1,−1(u)·v =µ(+)(u)v, (5.56)
s−1,1(u)·v =µ(−)(u)v. (5.57)
From the symmetry relation (5.53) we can obtain thatµi(u)s are not independent, every µi(u) can be expressed withµ1(u), µ2(u) and µ3(u). Using the defining equation (5.49) of X(so6,so5), the gl4 module L(λ1, λ2, λ3, λ4) also defines anX(so6,so5) module.
5.3 (SU(3),SO(3)) case
Now let us continue with the representation which come from the MPS. We consider the (SU(3), SO(3)) K-matrix as given by (4.14). From that representation we can derive the
following components in the complex basis, see also [30]
ψ1,1(s)(u) = 1−u−1Sz− 1
2u−2(s(s+ 1)−Sz(Sz+ 1)), (5.58)
ψ0,0(s)(u) = 1−u−2Sz2, (5.59)
ψ−1,−1(s) (u) = 1 +u−1Sz− 1
2u−2(s(s+ 1) +Sz(Sz+ 1)), (5.60) ψ0,−1(s) (u) =−2iu−1S−−iu−2SzS−, (5.61) ψ1,0(s)(u) = 2iu−1S−−iu−2S−Sz, (5.62)
ψ1,−1(s) (u) =u−2S−2, (5.63)
ψ−1,0(s) (u) = 2iu−1S++iu−2S+Sz, (5.64) ψ1,0(s)(u) =−2iu−1S++iu−2SzS+, (5.65)
ψ1,−1(s) (u) =u−2S+2, (5.66)
where Sx2+Sy2+Sz2=s(s+ 1). This is ak= 2s+ 1 dimensional irreducible representation of the twisted YangianY+(3). Let us denote itV(s). TheY+(3) highest weights ofV(s) are
µ1(u) = (1−su−1), (5.67)
µ0(u) = (1−s2u−2). (5.68)
From (5.44) and (5.45) we can see thatV(s) can be embedded into L(λ1, λ1, λ2) if λ1 =s andλ2 = 0 but L(s, s,0) is finite dimensional iff s∈N, therefore we only have chance to find connection between |Ψδi and|MPSki when kis odd.
For even k we have to use tensor product representations (see5.47and (5.48)). The representationL(λ1, λ1, λ2)⊗V(1/2) has highest weight
µ1(u) = (1 +λ2u−1)(1−λ1u−1)
1−1 2u−1
, (5.69)
µ0(u) = (1−λ21u−2)
1−1 4u−1
, (5.70)
we can see that V(s) can be embedded intoL(λ1, λ1, λ2)⊗V(1/2) if λ1=sandλ2 = 1/2.
5.3.1 Odd k= 2s+ 1
We have seen that the even and odd kcases must be treated differently. Let us start with the odd case. We can show thatV(s) is embedded intoL(s, s,0) where s∈Z+ for smalls.
These calculations can be found in appendix B. Using these explicit results we conjecture the embedding for generalsthe twisted YangianY+(3) acts onL(s, s,0)∼=V(s)⊕L(s, s,2) as
sij(u)·v1=L(s,s,0)ia (u)L(s,s,0)−j,−a(−u)v1=
= ψi,j(s)(u) X
0 L(s,s,2)ia (u)L(s,s,2)−j,−b(−u)
! w1 w2
!
, (5.71)
for ally∈Y+(3) where v∈L(s, s,0),w1 ∈V(s),w2 ∈L(s, s,2) ands∈Z>1. In (5.71) we used the following conjecture.
Conjecture 1. TheL(s, s,2)is an irrep of Y+(3)for all s >1.
Ratio of the overlaps In the following we will show that these results are consistent with the overlap formulas (4.12) and (4.13). So far, we used the convention (5.19) for the R-matrix but now we switch to the slightly different convention
R(u) =˜ uI+iP, (5.72)
and hence use rescaled matrices:
s˜i,j(u) =u2si,j(iu), (5.73)
˜ti,j(u) =uti,j(iu). (5.74)
Letρbe a representation of the the twisted Yangian and let us define the spectral parameter independent matrix
φij =ρ(sik(0))Ckj =ρ(si,−j(0)), (5.75) where C is the charge conjugation matrix. Furthermore, let us define the following state
|Ψi= X
i1,j1,...iL/2,jL/2
trAhφi1j1. . . φiL/2jL/2
i
i1, j1, . . . , iL/2, jL/2E. (5.76)
For the trivial representation Therefore the equation (5.71) connects|MPS2s+1i to the delta-state|Ψδias
|MPS2s+1i=T˜(s,s,0)(0)−T˜(s,s,2)(0)|Ψδi, (5.80) where
T˜(s,s,m)(u) = Tr0hL(s,s,m)01 (u). . .L(s,s,m)0L (u)i, (5.81) L˜(s,s,m)(u) = ˜L(s,s,m)i,j (u)⊗ei,j =uI−iEi,j(s,s,m)⊗ei,j =uI+iEi,j(−m,−s,−s)⊗ej,i, (5.82) where we used that Ei,j → −Ej,i is a Lie-algebra automorphism and it connects a represen-tation to its contra-gradient. Let us use another norepresen-tation:
L˜(s)(u) =u−is−1
2 +iEi,j(s,0,0)⊗ej,i, (5.83) T˜(s)(u) = Tr0
hL˜(s)01(u). . .L˜(s)0L(u)i. (5.84) From (5.82) and (5.83) we can see that
L˜(s,s,0)(u) = ˜L(s) and the ratio of the overlaps is equal to the difference of eigenvalues of transfer matrices
hMPS2s+1|ui
The eigenvalues of the transfer matrices can be written as [51]
T˜(s)(u) = Q0
Let us assume thatL, N0, N+ are even, then
We used the following conjecture.
Conjecture 2. TheL(s, s,3/2)⊗V(1/2) is an irrep of Y+(3) for all s∈Z++12. See appendixBfor the explanation.
Ratio of the overlaps In the following let us check that these results are consistent with the overlap formulas. The equation (5.94) connects|MPS2s+1i to|MPS2i as
|MPS2s+1i= 2
i L
T˜(s,s,1/2)(0)−T˜(s,s,3/2)(0)|MPS2i, (5.95)
where the factor (2/i)L comes from the prefactor of ψi,j(s)(u) in (5.94). From (5.82) and (5.83) we can see that
L˜(s,s,1/2)(u) = ˜L(m) and from the ratio of the overlaps we have to obtain that
hMPS2s+1|ui
5.4 (SO(6),SO(5)) case
The MPS can be built from the K-matrix ˜K =eab⊗ψab+e66⊗ψ66, whereψab andψ66 are given in equation (3.12) and (3.13). In the twisted Yangian language we use different normalization and basis. After the normalization
S(u) =−1
4u−3(1−u−1) ˜K(u), (5.105) and basis changing the S-matrix satisfies the reflection equation (5.52) and the symmetry relation (5.53). This is a highest weight representation and let us denote it by V(n). From the explicit forms
s3,3(u) = ˜g1(u) ˜G1G˜−1+ ˜g2(u)hG˜1,G˜−1
i+ ˜f(u), (5.106) s2,2(u) = ˜g1(u) ˜G2G˜−2+ ˜g2(u)hG˜2,G˜−2
i+ ˜f(u), (5.107) s1,1(u) = 1
2
g˜1(u) ˜G20+ ˜f(u) + ˜h(u), (5.108) s1,−1(u) = 1
2
g˜1(u) ˜G20+ ˜f(u)−˜h(u)=s−1,1(u), (5.109) we calculate the weights µ3(u),µ2(u),µ1(u) andµ(+)(u) =µ(−)(u). We used the notation
G˜±1= 1
√2(G1±iG2), (5.110)
G˜±2= 1
√
2(G3±iG4), (5.111)
G˜0=G5, (5.112)
and
˜g1(u) =−1
2u−2(1−u−2), (5.113)
˜g2(u) = 1
2u−1(1−u−2), (5.114)
f˜(u) = (1−u−1)(1 + C
4u−2), (5.115)
˜h(u) =−(1−u−1)(1 + 2u−1−C
4u−2). (5.116)
We can also calculate the highest weights of
SD(u) =Tso6(u)K(u) (Tso6)t(−u), (5.117) forgl4 moduleL(λ1, λ2, λ3, λ4). From the explicit calculation (see appendix C) we obtain thatV(n)∼=L(1 +n/2,1 +n/2,1 +n/2,1−n/2), i.e.
S(u) = 1−u−22
1− n2 + 12u−2Tso6(u)K(u) (Tso6)t(−u), (5.118) which implies that (with proper normalization of the states)
|MPSni= lim
u→0(2i)Lt(1)n (u)|Ψ0i, (5.119)
where t(a)n (u) is a solution of the Hirota equation (3.21). Let us use the z-functions
Using the tableau sum (3.27) the eigenvaluest(1)n (u) are then found to be t(1)n (u) = Q− −iu−n2 −1Q+ −iu+n2 + 1
With the present work we have reached a complete understanding of the integrability properties of a class of spin chain boundary states which among other things can be used for the calculation of one-point functions in domain wall versions ofN = 4 SYM theory. The boundary states in question take the form of matrix product states generated by matrices which are related to the generators of some irreducible representation of a Lie group. Matrix product states were in [5] characterized as integrable if annihilated by the odd charges of the underlying integrable spin chain. Furthermore, it was argued that this criterion being fulfilled would imply the existence of a boundary reflection matrix which together with the bulk R-matrix would fulfill a boundary Yang-Baxter relation which again should in principle make it possible to compute a number of quantum observables of the system in a closed form. The integrability criterion could immediately be shown to be fulfilled for two out of the three known relevant defect versions ofN = 4 SYM , namely the one dual to the D3-D5 probe brane system with flux and the one dual to theSO(5) symmetric D3-D7 probe brane system with non-vanishing instanton number [13], cf. table1, and a closed formula for the one-point functions of the former case could be found by numerical experiments [13].
A gap in the understanding was the apparent lack of a closed formula for one-point functions of the integrable D3-D7 probe brane set-up as well as the lack of an analytical method for the derivation of these one-point functions in all but the simplest cases for the D3-D5 set-up. With the present paper we have filled these gaps. First of all we have
obtained an understanding of the integrability properties for the D3-D7 probe brane set-up in a scattering picture by explicitly finding the appropriate boundary reflection matrix for any irreducible representation ofso(5) of the relevant type, cf. eqns. (3.9) and (3.12)-(3.13).
The corresponding boundary reflection matrix for the D3-D5 probe brane set-up was found in [8], se also [30]. Secondly we have explicitly derived the overlap formula (2.16) and (2.18) for the D3-D7 probe brane case by starting from a simple integrable one-site state and making use of the representation theory of twisted Yangians. We have also shown that a similar approach makes it possible to prove the formula for the D3-D5 case, earlier presented without proof [13], although for simplicity we completed the proof only for theSU(3) sector.
Interestingly, the derivation seems to be more involved for the supersymmetric D3-D5 case with SO(3) symmetric vevs than the non-supersymmetric D3-D7 with SO(5) symmetric vevs.
From the point of view of theoretical high energy physics, it would be extremely interesting if, using symmetries, one could bootstrap the integrable boundary reflection matrices to higher loop orders ofN = 4 SYM, as has been done for the bulk S-matrix [52], and derive the appropriate overlap formulas to all loop order. Again one would expect that a combination of the thermodynamical Bethe ansatz approach and the representation theory of twisted Yangians should be the correct way forward. A challenge is of course in the first place to take the present calculations beyond the scalar section of N = 4 SYM which is not closed at higher loop orders. An argument in favour of integrability of the defect systems at higher loop orders is that an exact expression for one-loop one-point functions ofSU(2) sub-sector of the D3-D5 probe-brane set-up has been found in [26] where also a possible asymptotic form of these was presented.
Another interesting extension of the present work is a possible generalization of the derivations to one-point functions of descendent states as these play an important role as data in the boundary conformal bootstrap program. One-point functions of descendent operators and their relevance for boundary conformal bootstrap was discussed for theSU(2) sub-sector of the D3-D5 probe brane set-up in [53].
Acknowledgments
MdL was supported by SFI, the Royal Society and the EPSRC for funding under grants UF160578, RGF\EA\181011, RGF\EA\180167 and 18/EPSRC/3590. C.K. was supported in part by DFF-FNU through the grant DFF-FNU 4002-00037. Furthermore, we thank Jan Ambjørn for giving us access to his computer system. The research of G.L. has received funding from the Hellenic Foundation for Research and Innovation (HFRI) and the General Secretariat for Research and Technology (GSRT), in the framework of the first post-doctoral researchers support, under grant agreement No. 2595. G.L. is thankful to the School of Mathematics of Trinity College Dublin for its hospitality and generous support. T.G. was supported in part by NKFIH grant K116505. B.P. was supported by the BME-Nanotechnology FIKP grant (BME FIKP-NAT), by the National Research Development and Innovation Office (NKFIH) (K-2016 grant no. 119204), by the János
Bolyai Research Scholarship of the Hungarian Academy of Sciences, and by the ÚNKP-19-4 New National Excellence Program of the Ministry for Innovation and Technology.
A Limiting formulas for the overlaps
As we have already mentioned the determinant formula (2.18) is strictly speaking valid only for even values of (n+ 1)N0/2. When (n+ 1)N0/2 is odd, one auxiliary Bethe root at each level vanishes and a 0/0 ambiguity arises. The indeterminate form is then treated in a standard way by noting that the terms in the square brackets of (2.18) become singular for q= 0,±1. In this case we obtain:
Eqn. (A.1) can also be in the following form:
Λn= 2L· The determinant formula (2.18) has been thoroughly checked to 50 digits of accuracy for many states of various lengthsL,N0 = 2, . . . ,10 andn= 1, . . . ,8, such that (n+ 1)N0/2 remains even. Itsu→0 limit that is given by (A.1)–(A.4) has also been checked for states of various lengthsL,N0 = 2,6,10 and even nsuch that (n+ 1)N0/2 is odd.
B Calculations with the twisted Yangian Y+(3)
In this section we investigate the embeddings V(s)⊆L(s, s,0) and V(s)⊆L(s, s,1/2)⊗ V(1/2) for the twisted Yangian Y+(3).
Table 2. The states ofL(2,2,0).
Sz = 2 |2,2,0i 1 |2,1,1i
0 |2,0,2i |1,2,1i
−1 |1,1,2i
−2 |0,2,2i
B.1 Odd k = 2s+ 1
s=1 For s= 1, the gl3 module L(1,1,0) has the same dimension asV(1) and the two have the same Y+(3) highest weights thereforeL(1,1,0)∼=V(1) as Y+(3) representation.
s=2 Fors= 2, the states of L(2,2,0) are shown in table 2where
|2,1,1i=E1,0|2,2,0i, (B.1)
|2,0,2i=E1,02 |2,2,0i, (B.2)
|1,2,1i=E0,−1E1,0|2,2,0i, (B.3)
|1,1,2i=E0,−1E1,02 |2,2,0i, (B.4)
|0,2,2i=E0,−12 E1,02 |2,2,0i. (B.5) We can see that theV(2) sub-space has to be built from the vectors|2,2,0i,|2,1,1i,|2,0,2i+
a|1,2,1i,|1,1,2i,|0,2,2i. The acan be calculated from the fact that s1,−1(u)· |2,2,0i ∈ V(2).
s1,−1(u)· |2,2,0i=t1,1(u)t1,−1(−u)· |2,2,0i+t1,0(u)t1,0(−u)· |2,2,0i+ +t1,−1(u)t1,1(−u)· |2,2,0i=
=−u−1(1 +u−1)E1,−1|2,2,0i −u−2|2,0,2i+u−1E1,−1|2,2,0i=
=u−2(|1,2,1i − |2,0,2i), (B.6)
therefore a=−1.
Let us define the sub-quotient L(2,2,0)\V(2) by W. The W is a one dimensional highest weight representation of Y+(3). Let us calculate the highest weights.
s11(u)· |2,0,2i=t1,1(u)t−1,−1(−u)· |2,0,2i= (1−4u−2)|2,2,0i, (B.7) s00(u)· |1,2,1i=t0,0(u)t0,0(−u)· |1,2,1i= (1−4u−2)|2,2,0i. (B.8) Therefore
µ1(u) = (1−4u−2), (B.9)
µ0(u) = (1−4u−2), (B.10)
Table 3. The states ofL(3,3,0).
Sz = 3 |3,3,0i 2 |3,2,1i
1 |3,1,2i |2,3,1i 0 |3,0,3i |2,2,2i
−1 |2,1,3i |1,3,2i
−2 |1,2,3i
−3 |0,3,3i
which are the highest weights of one dimensional irrep L(2,2,2). Hence, for L(2,2,0)∼= V(2)⊕L(2,2,2) the action ofY+(3) reads as
sij(u)·v=L(2,2,0)ia (u)L(2,2,0)−j,−a(−u)v= ψ(2)ij (u) X
0 L(2,2,2)ia (u)L(2,2,2)−j,−a(−u)
! w1
w2
!
, (B.11) where v∈L(2,2,0),w1 ∈V(2) andw2 ∈L(2,2,2).
s=3 Let us continue with s= 3. The table 3 shows the states of L(3,3,0). We can see that theV(3) sub-space has to be built from the vectors|3,3,0i,|3,2,1i,|3,1,2i+a1|2,3,1i,
|3,0,3i+a2|2,2,2i,|2,1,3i+a3|1,3,2i,|1,2,3i,|0,3,3i. Thea1 can be calculated from the fact that s1,−1(u)· |3,3,0i ∈V(3).
s1,−1(u)· |3,3,0i=t1,1(u)t1,−1(−u)· |3,3,0i+t1,0(u)t1,0(−u)· |3,3,0i+ +t1,−1(u)t1,1(−u)· |3,3,0i=
=−u−1(1 +u−1)E1,−1|3,3,0i −u−2|3,1,2i+u−1E1,−1|3,3,0i=
=u−2(|2,3,1i − |3,1,2i), (B.12)
therefore a1 =−1.
Let us define the sub-quotient L(3,3,0)\V(3) by W. The W is a three dimensional highest weight representation of Y+(3). Let us calculate the highest weights.
s11(u)· |3,1,2i=t1,1(u)t−1,−1(−u)· |3,1,2i= (1 + 2u−1)(1−3u−1)|3,1,2i, (B.13) s00(u)· |2,3,1i=t0,0(u)t0,0(−u)· |2,3,1i= (1−9u−2)|2,3,1i. (B.14) Therefore
µ1(u) = (1 + 2u−1)(1−3u−1), (B.15)
µ0(u) = (1−9u−2). (B.16)
Table 4. The states ofL(3,3,2).
Sz = 1 |3,3,2i 0 |3,2,3i
−1 |2,3,3i
Table 5. The states ofL(4,4,2).
Sz = 2 |4,4,2i 1 |4,3,3i
0 |4,2,4i |3,4,3i
−1 |3,1,4i
−2 |2,4,4i
We can see that these are the highest weights of L(3,3,2). TheL(3,3,2) is an Y+(3) irrep (see table 4) i.e. forL(3,3,0)∼=V(3)⊕L(3,3,2) the action of Y+(3) reads as
sij(u)·v=L(3,3,0)ia (u)L(3,3,0)−j,−a(−u)v= ψ(3)ij (u) X
0 L(3,3,2)ia (u)L(3,3,2)−j,−a(−u)
! w1
w2
!
, (B.17) where v∈L(3,3,0),w1 ∈V(3) andw2 ∈L(3,3,2).
s=4 Similarly, for L(4,4,0) we can define the sub-quotient W = L(4,4,0)\V(4). The highest weights of it are the following
µ1(u) = (1 + 2u−1)(1−4u−1), (B.18)
µ0(u) = (1−16u−2). (B.19)
These are also the highest weights ofL(4,4,2).We can show that this is anY+(3) irrep. The states are shown in table5. We only have to show that the subspacesij(u)· |4,4,2icontains the two dimensional subspace span ({|4,2,4i,|3,4,3i}). This can be done as follows
s1,0(u)· |4,3,3i=t1,1(u)t0,−1(−u)· |4,3,3i+t1,0(u)t0,0(−u)· |4,3,3i+ +t1,−1(u)t0,1(−u)· |4,3,3i=
=−u−1(1 + 3u−1)|3,4,3i+u−1(1−3u−1)|4,2,4i −2u−2Λ1,−1|4,4,2i=
=u−1(|4,2,4i − |3,4,3i)−u−2(3|4,2,4i+|3,4,3i). (B.20) SinceL(4,4,2) is an irrep with the same highest weights as W thereforeW ∼=L(4,4,2) asY+(3) representations i.e. for the vector space decompositionL(4,4,0)∼=V(4)⊕L(4,4,2)
Table 6. The states ofL(3/2,3/2,1/2)⊗V(1/2).
s=3/2 Let us start with s= 3/2. The states are shown in table 6. Let us define the sub-quotient L(3/2,3/2,1/2)⊗V(1/2)\V(3/2) byW. The W is a two dimensional highest weight representation of Y+(3). Let us calculate the highest weights.
s11(u)·
Therefore the highest weights are µ1(u) =
Table 7. The states ofL(5/2,5/2,1/2)⊗V(1/2). a six dimensional highest weight representation of Y+(3). We calculate the highest weights:
s11(u)·
Therefore the highest weights are µ1(u) =
Table 8. The states ofL(5/2,5/2,3/2)⊗V(1/2).
We can see that these are the highest weights of L(5/2,5/2,3/2)⊗V(1/2). We can show that this is an Y+(3) irrep. The states are shown in table8.
C Calculations with the twisted Yangian X(so6,so5) From the explicit forms
s3,3(u) = ˜g1(u) ˜G1G˜−1+ ˜g2(u)hG˜1,G˜−1
i+ ˜f(u), (C.1) s2,2(u) = ˜g1(u) ˜G2G˜−2+ ˜g2(u)hG˜2,G˜−2
i+ ˜f(u), (C.2) s1,1(u) = 1
2
g˜1(u) ˜G20+ ˜f(u) + ˜h(u), (C.3) s1,−1(u) = 1
2
g˜1(u) ˜G20+ ˜f(u)−˜h(u)=s−1,1(u), (C.4) we can calculate the weights µ3(u),µ2(u),µ1(u) and µ(+)(u) =µ(−)(u).
Let us define the so5 generators
Fij = 1
4[Gi, G−j], (C.5)
for which
hFij,G˜ki=δjkG˜i−δ−i,kG˜−j. (C.6) TheFijs form a h.w. representation ofso5 i.e. there exist a vectorv for which (λ1, λ2) = (−n2,−n2). From (C.6) we can see that ˜G1 and ˜G2 increase the weights and ˜G−1 and ˜G−2
decrease the weights, therefore
G−1·v=G−2·v= 0. (C.7)
Using this we can calculateµ3 andµ2 as s3,3(u)·v=f˜(u) + 4˜g2(u)F1,1
·v=f˜(u)−2n˜g2(u)v, (C.8) s2,2(u)·v=f˜(u) + 4˜g2(u)F2,2
·v=f˜(u)−2n˜g2(u)v. (C.9) Forµ1 and µ(+) we have to calculate how ˜G20 act onv. This can be done using (3.14)
G˜20·v=Cv−G˜1G˜−1+ ˜G−1G˜1+ ˜G2G˜−2+ ˜G−2G˜2
·v=
=Cv+ 4 (F11+F22)·v= (C−4n)v=n2v. (C.10) Using this we obtain that
s1,1(u)·v= 1 2
n2g˜1(u) + ˜f(u) + ˜h(u)·v, (C.11) s2,2(u)·v= 1
2
n2g˜1(u) + ˜f(u)−˜h(u)·v. (C.12) Therefore
µ3(u) =µ2(u) =1−u−1
1−n 2u−1
2
, (C.13)
µ1(u) =−u−11−u−1
1−n 2u−1
2
, (C.14)
µ(+)(u) =µ(−)(u) =1−u−2 1−n2 4 u−2
!
. (C.15)
Now let us try to obtain this S-matrix in the form
SD(u) =Tso6(u)K(u) (Tso6)t(−u), (C.16) using somegl4 moduleL(λ1, λ2, λ3, λ4). Theso5is embedded ingl4 andX(so6,so5). Our S-matrix andL(λ1, λ2, λ3, λ4) have highest weights (−n2,−n2) and−λ1+λ2−λ2 3−λ4,−λ1−λ2+λ2 3−λ4 asso5 modules. Therefore thegl4 highest weights have to be (n+c, a, a, c). Matching the dimensions of the representation, only two options remained.
1. (n+c, n+c, n+c, c) or 2. (n+c, c, c, c).
Let us calculate the X(so6,so5) highest weights of the first one.
sD3,3(u)·v= u
u+ 1tso3,36(u)tso−3,−36 (u)·v= (C.17)
= u
u+ 1tgl334(u)tgl444(u−1)tgl114(−u)tgl224(−u−1)·v= (C.18)
= (u+n+c)(u+c−1)(u−n−c)(u−n−c+ 1)
u(u+ 1)2(u−1) ·v, (C.19)
where we used that tsoi,j6(u)·v = tgli,j4(u)·v = 0 for i < j. In an analogous way we can calculate the other highest weights and the result is
µD3(u) =µD2(u) =
u2−(n+c)2(u+c−1)(u−n−c+ 1)
u(u+ 1)2(u−1) , (C.20) µD1(u) =−
u2−(n+c)2(u+c−1)(u−n−c+ 1)
u2(u+ 1)2(u−1) , (C.21)
µD(+)(u) =
u2−(n+c)2 u2−(n+c−1)2
u2(u2−1) , (C.22)
µD(−)(u) =
u2−(n+c)2 u2−(c−1)2
u2(u2−1) . (C.23)
The equation (C.15) implies that µD(+)(u) has to equal toµD(−)(u) therefore c= 1−n
2. (C.24)
After substitution µD3 (u) =µD2(u) =
1− n2 + 12u−2 1−n2u−12 (1 +u−1)2(1−u−1) =
1− n2 + 12u−2 (1−u−2)2 µ3(u),
(C.25) µD1 (u) =−u−1
1− n2 + 12u−2 1− n2u−12 (1 +u−1)2(1−u−1) =
1− n2 + 12u−2
(1−u−2)2 µ1(u), (C.26) µD(+)(u) =µD(−)(u) =
1− n2 + 12u−2 1− n42u−2
(1−u−2) =
1− n2 + 12u−2
(1−u−2)2 µ(+)(u).
(C.27)
Therefore we obtain that
S(u) = 1−u−22
1− n2 + 12u−2Tso6(u)K(u) (Tso6)t(−u). (C.28) References
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